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3 years ago

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A 0.125 kg hockey puck moving at 20.0 m/s is caught and held by an 85.0 kg goalkeeper at rest. After catching the puck, with wha

t velocity does the goalkeeper slide across the ice?
A.
0.0351 m/s

B.
0.0294 m/s

C.
0.0320 m/s

D.
0.0263 m/s

1 answer:

jenyasd209 [6]3 years ago

4 0

Answer:

a

Explanation:cuz i saiad so

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A 86g ball is dropped vertically to the floor from a height of 2.87m and bounces to a height of 1.28. What is the magnitude of t

irga5000 [103]

Answer:

The impulse received by the ball from the floor during the bounce is approximately 1.11329438 m·kg/s

Explanation:

The given mass of the ball, m = 86 g = 0.089 kg

The height from which the ball is dropped, H = 2.87 m

The height to which the ball bounces, h = 1.28 m

Mathematically, we have;

Δp = F·Δt

Where;

Δp = The change in momentum = m·Δv

F = The applied force

Δt = The time of contact with the force

The velocity of the ball just before it touches the ground, v₁ = -√(2·g·H)

The velocity with which the ball leaves, v₂ = √(2·g·h)

The change in momentum, Δp = m·(v₂ - v₁)

∴ Δp = m·(√(2·g·h) - (-√(2·g·H))) = m·(√(2·g·h) +√(2·g·H) )

The impulse, Δp, received by the ball from the floor during the bounce is given as follows;

Δp = 0.089 kg × (√(2 × 9.8 m/s² × 1.28 m) + √(2 × 9.8 m/s² × 2.87 m)) ≈ 1.11329438 m·kg/s

The impulse received by the ball from the floor during the bounce, Δp ≈ 1.11329438 m·kg/s

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3 years ago

A bomb, originally sitting at rest, explodes and during the

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Answer:

opposite

Explanation:

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3 years ago

What is the volume V of a sample of 4.00 mol of copper? The atomic mass of copper (Cu) is 63.5 g/mol, and the density of copper

rusak2 [61]

Answer : The volume of a sample of 4.00 mol of copper is $28.5cm^3$

Explanation :

First we have to calculate the mass of copper.

$\text{ Mass of copper}=\text{ Moles of copper}\times \text{ Molar mass of copper}$

$\text{ Mass of copper}=(4.00moles)\times (63.5g/mole)=254g$

Now we have to calculate the volume of copper.

Formula used :

$Density=\frac{Mass}{Volume}$

Now put all the given values in this formula, we get:

$8.92\times 10^3kg/m^3=\frac{254g}{Volume}$

$Volume=\frac{254g}{8.92\times 10^3kg/m^3}=2.85\times 10^{-2}L=2.85\times 10^{-2}\times 10^3cm^3=28.5cm^3$

Conversion used :

$1kg/m^3=1g/L\\\\1L=10^3cm^3$

Therefore, the volume of a sample of 4.00 mol of copper is $28.5cm^3$

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If the force is 4 newtons between two charged spheres separated by 3 centimeters, calculate the force between the same spheres s

il63 [147K]

Answer:

1 Newton

Explanation:

F=9*10^9*q0q1/r^2]]

F=9*10^9*(q0q1)/ r^2

r=3cm

F=4N

F=9*10^9*(q0q1)/3^2

4=9*10^9*(q0q1)/9

4=10^9 q0q1

q0q1=4/10^9

q0q1=4*10^-9

To calculate the force between the forces at a distance of 6 cm

F=9*10^9*(q0q1)/ r^2

=9*10^9*(4*10^-9)/6^2

=9*10^9*(4*10^-9)/36

=10^9*4*10^-9/4

=10^9*10^-9

=1 Newton

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Salt + water because salt dissolved in water

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